Question

Com a fórmula de Bhaskara, resolva as seguintes equações:

Answer 1

a)
$-x^2 + 8x+9=0$

$a = -1 \\ b = 8 \\ c = 9$

$x = \frac{-b\pm \sqrt{b^2-4.a.c} }{2.a} \\ \\ x = \frac{-8\pm \sqrt{8^2-4.(-1).9}}{2.(-1)} \\ \\ x = \frac{-8\pm \sqrt{64-(-36)} }{-2} \\ \\ x = \frac{-8\pm \sqrt{100} }{-2} \\ \\ x = \frac{-8\pm10}{-2} \\ \\ x' = \frac{-8+10}{-2} \rightarrow x' = \frac{2}{-2} \rightarrow x' = -1 \\ \\ x" = \frac{-8-10}{-2} \rightarrow x" = \frac{-18}{-2} \rightarrow x" = 9$


b)

$x^2+9x+8 = 0$

$a = 1 \\ b = 9 \\ c = 8$

$\Delta = b^2-4.a.c \\ \Delta = 9^2 - 4.1.8 \\ \Delta = 81 - 32 \\ \Delta = 49 \\ \\ x = \frac{-b\pm \sqrt{b^2-4.a.c} }{2.a} \\ \\ x = \frac{-9\pm \sqrt{49} }{2.1} \\ \\ x = \frac{-9\pm7}{2} \\ \\ x' = \frac{-9+7}{2} \rightarrow x' = \frac{-2}{2} \rightarrow x' = -1 \\ \\ x" = \frac{-9-7}{2} \rightarrow x" = \frac{-16}{2} \rightarrow x" = -8$


c)

$x^2+10x+24=-1 \\ x^2 + 10x +25=0 \\ \\ a = 1 \\ b=10 \\ c = 25 \\ \\ \Delta = b^2-4.a.c \\ \Delta = 10^2 - 4.1.25 \\ \Delta = 100-100 \\ \Delta=0 \\ \\ x = \frac{-b\pm \sqrt{b^2-4.a.c} }{2.a} \\ \\ x = \frac{-10\pm \sqrt{0} }{2.1} \\ \\ x = \frac{-10}{2} \\ \\ x = -5$


d)

$3x^2 - 2x = 0$

$a = 3 \\ b = -2 \\ c = 0$

$\Delta = b^2-4.a.c \\ \Delta = (-2)^2 - 4.3.0 \\ \Delta = 4 \\ \\ x = \frac{-b\pm \sqrt{b^2-4.a.c} }{2.a} \\ \\ x = \frac{-(-2)\pm \sqrt{4} }{2.3} \\ \\ x = \frac{2\pm2}{6} \\ \\ x' = \frac{2+2}{6} \rightarrow x' = \frac{4}{6} \rightarrow x' = \frac{2}{3} \\ \\ x" = \frac{2-2}{6} \rightarrow x" = 0$


e)

$x^2 - 2 \sqrt{2} x+1 = 0 \\ \\ a = 1 \\ b =- 2 \sqrt{2} \\ c = 1 \\ \\ \Delta = b^2-4.a.c \\ \Delta = (-2 \sqrt{2} )^2 - 4.1.1 \\ \Delta = 8 - 4 \\ \Delta = 4 \\ \\ x = \frac{-b\pm \sqrt{b^2-4.a.c} }{2.1} \\ \\ x = \frac{-(-2 \sqrt{2})\pm \sqrt{4} }{2.1} \\ \\ x = \frac{2 \sqrt{2}\pm2 }{2} \\ \\ x' = \frac{2 \sqrt{2}+2 }{2} \rightarrow x' = \sqrt{2} + 1 \\ \\ x" = \frac{2 \sqrt{2}-2 }{2} \rightarrow x" = \sqrt{2} - 1$


e)

$4x^2 - 9 = 0 \\ \\ a = 4 \\ b = 0 \\ c = -9 \\ \\ \Delta = b^2-4.a.c \\ \Delta = 0-4.4.(-9) \\ \Delta = 144 \\ \\ x = \frac{-b\pm \sqrt{b^2-4.a.c} }{2.} \\ \\ x = \frac{0\pm \sqrt{144} }{2.4} \\ \\ x = \frac{\pm12}{8} \\ \\ x' = \frac{12}{8} \rightarrow x' = \frac{3}{2} \\ \\ x" = \frac{-12}{8} \rightarrow x" = - \frac{3}{2}$


Ok?